What is the integral of (x^2)/(x^6+2x^3+3) ?

The integral of (x^2)/(x^6+2x^3+3) is 1/(3sqrt(5))arctan((x^3)/(sqrt(5)))+C

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integral of (x^2)/(x^6+2x^3+3)

\int \frac{x ^{2}}{x ^{6} + 2 x ^{3} + 3} d x

\frac{1}{3 5} arctan (\frac{x ^{3}}{5}) + C

Solution steps

\int \frac{x ^{2}}{x ^{6} + 2 x ^{3} + 3} d x

Apply u-substitution

: \int \frac{1}{6 u ( 2 u + u + 3 )} d u

= \int \frac{1}{6 u ( 2 u + u + 3 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot \int \frac{1}{u ( 2 u + u + 3 )} d u

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{u ( 2 u + u + 3 )} = u^{- \frac{1}{2}} \frac{1}{( 2 + u + 3 )}

= \frac{1}{6} \cdot \int u^{- \frac{1}{2}} \frac{1}{2 + u + 3} d u

Simplify

u^{- \frac{1}{2}} \frac{1}{2 + u + 3} : \frac{1}{u ( u + 5 )}

= \frac{1}{6} \cdot \int \frac{1}{u ( u + 5 )} d u

Apply u-substitution

: \int \frac{2}{v ^{2} + 5} d v

= \frac{1}{6} \cdot \int \frac{2}{v ^{2} + 5} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot 2 \cdot \int \frac{1}{v ^{2} + 5} d v

Apply Integral Substitution

= \frac{1}{6} \cdot 2 \cdot \int \frac{1}{5 ( w ^{2} + 1 )} d w

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot 2 \cdot \frac{1}{5} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

\int \frac{1}{w ^{2} + 1} d w = arctan (w)

= \frac{1}{6} \cdot 2 \cdot \frac{1}{5} arctan (w)

Substitute back

= \frac{1}{6} \cdot 2 \cdot \frac{1}{5} arctan (\frac{x ^{6}}{5})

Simplify

\frac{1}{6} \cdot 2 \cdot \frac{1}{5} arctan (\frac{x ^{6}}{5}) : \frac{1}{3 5} arctan (\frac{x ^{3}}{5})

= \frac{1}{3 5} arctan (\frac{x ^{3}}{5})

Add a constant to the solution

= \frac{1}{3 5} arctan (\frac{x ^{3}}{5}) + C

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What is the integral of (x^2)/(x^6+2x^3+3) ?
The integral of (x^2)/(x^6+2x^3+3) is 1/(3sqrt(5))arctan((x^3)/(sqrt(5)))+C